如何在C＃中使用Linq计数与group by

Question

i have a datatable in which look like this

basically i want it to be group by column featurename(distinct) in which it should sum in effort and complete column
specify all the featureid,featurename comma seprated
count of the featureid
assigned to comman seprated
now i want datatable to be look like this

don't know how to use the count code

var result5= dtTaskandBugs.AsEnumerable().GroupBy(x => x["Functional Area"])
                               .Select(item => new
                               {
                                   FunctionalArea = item.Key,

                                   Completedsum = item.Sum(y => Convert.ToDecimal(y["Completed"])),
                                   effortsum = item.Sum(z => Convert.ToDecimal(z["effort"])),
                                   storyids = string.Join(",", item.Select(a => a["Storyid"]).Distinct()),
                                   storiesename= string.Join(",", item.Select(b => b["StoryName"]).Distinct()),
                                   Featureid = string.Join(",", item.Select(c => c["Featureid"]).Distinct()),
                                   Featurename= string.Join(",", item.Select(d => d["FeatureName"]).Distinct()),

                               });

Answer 1

With Count() you will get count of grouped value. Try with Linq like this:

from s in dtTaskandBugs.AsEnumerable()
                        group s by s.Field<string>("Functional Area")
                            into grp
                            orderby grp.Key
                            select new { 
                                   FunctionalArea = grp.Key,
                                   Completedsum = grp.Sum(y => Convert.ToDecimal(y["Completed"])),
                                   effortsum = grp.Sum(z => Convert.ToDecimal(z["effort"])),
                                   storyids = string.Join(",", grp.Select(a => a["Storyid"]).Distinct()),
                                   storiesename= string.Join(",", grp.Select(b => b["StoryName"]).Distinct()),
                                   Featureid = string.Join(",", grp.Select(c => c["Featureid"]).Distinct()),
                                   Featurename= string.Join(",", grp.Select(d => d["FeatureName"]).Distinct()),
                                   Count = grp.Count()          // <----------------
 };

Answer 2

只需编写item.count() ，您将获得分组值的计数，